The question at hand is basically, given positive $x_1, x_2, ... $ such that they sum to one (they are an infinite sequence of positive numbers), and given $y_1, y_2, ... $ as a sequence of distinct points in $\mathbb{R}$, find a measurable function $f:[0,1]\rightarrow \mathbb{R} $ such that the pushforward of the Lebesgue measure, $\mu = f_*m$, satisfies $\mu(E) \geq x_i$ for all $y_i \in E$.

How would I approach this problem? Does anyone know of a working function? Help would be greatly appreciated!

1 Answer
1

If I understood the question correctly, then I assume you require $\mu$ to be such that $\mu(E)\geq x_{i}$ for all those $i\in \mathbb{N}$ for which $y_{i}\in E$.

In particular we will want $m(f^{-1}(\{y_{i}\}))\geq x_{i}$ for every $i\in \mathbb{N}$. We use the fact that $\sum_{i=1}^{\infty}x_{i}=1$ and we choose $x_{0}:=0$. Let $f:[0,1]\to \mathbb{R}$ be such that:

In other words, we set $f(x)=y_{i}$ if $x\in [\sum_{n=0}^{i-1}x_{n},\sum_{n=0}^{i}x_{n})$. Hence the preimage of each $y_{i}$ is the interval $[\sum_{n=0}^{i-1}x_{n},\sum_{n=0}^{i}x_{n})$ which has Lebesgue measure of $x_{i}$, which guarantees that the condition that you needed holds. Hence $\mu(E)=\sum_{y_{i}\in E}x_{i}\geq x_{i}$ for those $i$ when $y_{i}\in E$.

The preimage of each open subset $U\subset \mathbb{R}$ is a Borel set of $[0,1]$, so $f$ is a measurable function.